I am having a hard time understanding difference between parallel adder and serial adder when adding two binary numbers. Wikipedia does not have any information, and google image search is bringing me almost nothing. Can anyone help me here?


1 Answer 1


If you want to add two N-bit numbers, a parallel adder will use N copies of a full-adder circuit, and it will produce a result in one clock cycle.

On the other hand, a serial adder will use just one full-adder circuit, but it will require N clocks to produce the result.

The tradeoff is circuit complexity (and power consumption) vs. time.

Some examples: A desktop PC uses a parallel adder, because it wants to be able to do as much arithmetic as possible in a given amount of time, and there is no real restriction on how much power it can consume. On the other hand, a handheld calculator only needs to be as fast as the human operating it, but it needs to last a long time on a small battery, so it uses a serial adder.

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    \$\begingroup\$ "a parallel adder ... will produce a result in one clock cycle." except for small but irritating problem of the (ripple) carry (or it must be a very long clock cycle). \$\endgroup\$ Mar 23, 2014 at 17:42
  • \$\begingroup\$ @WoutervanOoijen: No, carry look-ahead logic makes it (near-)constant time. \$\endgroup\$
    – Dave Tweed
    Mar 23, 2014 at 22:56
  • \$\begingroup\$ Yep, but it that case your N-bit adder will be (much) more than just N 1-bit-full-adders! \$\endgroup\$ Mar 24, 2014 at 6:43
  • \$\begingroup\$ @WoutervanOoijen: I can't tell ... are you making a point that's actually relevant to the original question? \$\endgroup\$
    – Dave Tweed
    Mar 24, 2014 at 11:23
  • \$\begingroup\$ Your first line states/implies that an N-bit parallel adder is JUST N-copies of a full-adder and PRODUCES A RESULT IN ONE CLOCK CYCLE. In contrast with a serial adder, who needs N clocks. That comparison is misleading because a serial adder needs no carry-look-ahead logic, but a parallel adder does (for a compareable clock speed), and this is is significant part of a parallel adder. \$\endgroup\$ Mar 24, 2014 at 14:01

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